# I No slip condition in an ideal fluid- Perpendicular pressure

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1. Oct 2, 2016

### Timtam

The no slip condition has been described as the adhesion of a fluid to a solid surface setting the relative fluid velocity to zero - cohesion (viscous stress) between fluid elements spreads evenly the velocity gradient through the boundary to the free stream.
This also infers that the pressure at the boundary would still be static (zero velocity) and by Pascals purely perpendicular to the surface

1. The problem statement, all variables and given/known data
In an ideal fluid (Without the presence of adhesion,cohesion and viscous stress) would a thin layer of fluid due purely to its kinetic interactions with the adjacent solid - still experience zero fluid velocity, meaning that the fluid pressure still be static and act perpendicular near the fixed surface.

2. Relevant equations

3. The attempt at a solution
I believe it would - Digging a bit deeper into why pressure is always perpendicular at a boundary even in the presence of a Dynamic Pressure (organised momentum) I came across a slightly different explanation than adhesion.
I believe the explanation is Maxwell Specular reflection - in short the momentum provided to the solid surface is absorbed by the wall via its only available degrees of freedom -rotational.
At the same time the rotational degrees of freedom of the wall provide the momentum to the translational degrees of freedom of the fluid molecule.
As there is no relationship between the original linear momentum and the final linear momentum of the fluid. The final momentum will be across all vectors away from the wall - random static pressure with all tangential components cancelled .
This specular reflection seems to be a robust case that the no slip condition would still exist in a ideal fluid
its only the even velocity gradient that wouldn't exist

2. Oct 3, 2016

3. Oct 3, 2016

### boneh3ad

The orthodoxy is that the no-slip condition does not exist in an inviscid fluid. I had trouble following his reasoning either because it is currently early in the morning or else because he kept mixing/misusing terminology. Probably a combination of the two.

4. Oct 3, 2016

### boneh3ad

So let me dissect this a bit to see if I can figure out what is going on here...

"Static" pressure is not defined as the pressure when a fluid is brought to rest. Static pressure is simply the thermodynamic pressure in a flow, i.e. the pressure felt by a surface or an adjacent control volume. Stagnation (or total) pressure is the pressure achieved when a flow is brought isentropically to rest. However, boundary layers are dissipative, and as a result of viscous dissipation, the total pressure actually decreases as one approaches the wall. As it turns out, the wall-normal static pressure gradient is effectively zero, so the pressure at the wall is generally equal to the pressure at the edge of the boundary layer. This phenomenon is the reason why inviscid flows can be a useful tool for modeling the pressure forces in viscous flows.

Further, pressure is isotropic and purely a normal stress. It cannot exert a shear stress on a surface, regardless of viscosity. This is true whether a boundary layer forms due to viscosity or not. However, perhaps you are using the term "pressure" loosely here? Do you instead mean "stress"? If so, then you have it backward. The stress tensor for a Newtonian fluid is defined as
$$\sigma_{ij} = -p\delta_{ij} + \mu\left( \dfrac{\partial v_i}{\partial x_j} + \dfrac{\partial v_j}{\partial x_i} \right).$$
In the absence of viscosity, $\mu = 0$ and the stresses are simply
$$\sigma_{ij} = -p\delta_{ij},$$
which is a diagonal tensor that acts only normal to a surface. For a viscous fluid, you still have the diagonal pressure terms, but you also have the viscous term,
$$\mu\left( \dfrac{\partial v_i}{\partial x_j} + \dfrac{\partial v_j}{\partial x_i} \right),$$
which contains nonzero off-diagonal elements based on the velocity gradient. So, in fact, having viscosity, and therefore a velocity gradient, means that the overall stress does not act purely normal to a surface. This is where viscous drag arises, and it is why inviscid models can predict things like lift and drag due to pressure with good accuracy, but can't even remotely predict total drag on an object since they don't handle viscosity.

Due to the above, the highlighted statement is a non sequitur. Stagnation at the wall does not enforce pressure as a normal stress. That occurs by default.

You are correct that the fact that pressure acts normal to a surface can be explained with specular reflections. However, this does not mean that the no slip conditions exists in an inviscid fluid. The pressure still acts normal to the surface but the fluid would also slip along the surface. I think the problem you are experiencing here is that you are equating the normal action of pressure with the stagnation of the flow at the wall, which are not equivalent phenomena. So, while you have correctly reasoned that the pressure acts normal to a surface whether the flow is viscous or not, this says nothing about whether the no-slip condition exists.

Hopefully that helps.

5. Oct 3, 2016

### Staff: Mentor

To add to what Boneh3ad said, if you are talking about the force per unit area acting on a surface, you are talking about stress, and not pressure. Only in a case of hydrostatic equilibrium is the pressure the same as the stress. In general, if the fluid is deforming (and not at hydrostatic equilibrium), the state of stress of the fluid is resolved mathematically into two separate parts: an isotropic pressure (which acts equally in all directions and results in forces that are perpendicular to all surfaces) and a viscous part (which is typically not perpendicular to the surface). If the fluid is in hydrostatic equilibrium, the viscous part is zero, and the only part left is the isotropic pressure.

So you are correct in your assessment of the state of stress when viscous flow is occurring. But, make sure you recognize that the term "pressure" is not descriptive of what is happening. The correct term to use is "stress," and this is the linear combination of the isotropic pressure and the viscous contribution to the stress.

6. Oct 3, 2016

### olivermsun

I'm a little confused myself, but I think the OP's example is talking about an ideal, or perfect, fluid with only an isotropic pressure and no viscous stresses, and asking whether kinetic (elastic?) interactions are enough to cause the tangential velocity to go to zero at the boundary.

7. Oct 3, 2016

### boneh3ad

And the answer to that is "no". I think he is convinced that the pressure being isotropic at the wall is a result of viscosity and the no-slip condition, and therefore is using that to show that, since the pressure at the wall is still isotropic in an ideal fluid flow, the velocity must still be zero. The problem is that the first premise is incorrect, and the pressure is simply always isotropic, and therefore says nothing about whether the no-slip conditions applies or not.

8. Oct 3, 2016

### olivermsun

I understand that the answer is "no," but I assume he has to explain his answer based on the way the question was asked.

Seems his concept is backwards, then. But anyway, one should be able to ignore the terminology of viscosity (since it given to be absent) and just consider kinetic interactions. If molecules of the fluid are bouncing "specularly" off the wall, and there is absolutely no non-normal momentum change, then it seems that the question is answered.

9. Oct 4, 2016

### Timtam

@boneh3ad @Bystander @Chestermiller @olivermsun Thanks for all your reply's.
I'm sorry if my phrasing is confusing, instead of replying to all the different points, I will attempt to simplify my question with the help of some diagrams.

Firstly I don't really care about the no slip condition or viscosity - I care about understanding how I can get from a solid surface any vector of momentum-Thin Dotted Blue (averaged to perpendicular -Thick dotted Blue ) regardless of the original momentum vector. Solid Blue Line

The ideal case is just the best way I could isolate the cause of this.

I am approaching this from Momentum Conservation so
is most correct except for I understand that these interactions are completely inelastic

To explain the difference I begin with Elastic collisions

Elastic Collison's -Dynamic Case Ideal Fluid
For a single molecule with an original momentum approaching the solid surface- it has a perpendicular and tangential component

In an elastic collision with the wall I would expect that the angle of incidence and reflection equal

This means that the tangential component F.SinX is unchanged and perpendicular F.CosX is reversed (negative) as a result .

This change in only perpendicular velocity agrees with a purely perpendicular force but also indicates slip to me, as FSinx+ is unchanged

Elastic Collison's -Static Case Ideal Fluid
In an ideal fluid static fluid I can see on average (with two molecules red and blue randomly following opposite paths) I can see that the tangents cancel and we are left again with just a perpendicular force.

So in both the ordered and random momentum elastic scenarios we end up with a purely perpendicular force.

Inelastic collisions
The problem is that I understand from Maxwells theory that these molecular collisions are completely inelastic and actually only add to the wall, and receive from the wall , rotational or vibrational momentum -Heat . The exchange of only heat accounts for the completely random resultant momemtum vectors.

Applied to the static case
As the original momentum is already random I see that has no effect on the static case
Applied to the dynamic case
In the dynamic case as the resultant vector is irrelevant to original we will no ordered momentum remaining FsinX=0 we have completely disordered momentum vectors.

This seems to be an enforcement of the no slip based purely on kinetic interactions, without the need for viscosity.

Again I don't really care about the no slip condition. I care about understanding this random resultant momentum vectors from the originally ordered momentum vectors in the Dynamic case .

Does this help simplify my question ?

10. Oct 4, 2016

### Bystander

11. Oct 4, 2016

### A.T.

I don't understand this definition of "completely inelastic" at the molecular level. Are the molecules stuck together after collision?

How does viscosity work on the molecular level, if not by kinetic interactions?

It's the gradient of the flow velocity towards the wall.

12. Oct 4, 2016

### Timtam

Yes I understand that to be the case . The fluid molecule imparts all its momentum to the molecules in the wall, in effect coming to rest . the wall then imparts unrelated momentum back on to the fluid molecule . It needn't be thought of as sequential rather simultaneous.

This inelasticity is what I understand allows the completely random resultant vectors to occur

Again sorry for the confusing language, I meant by the forces of adhesion and cohesion which I understand to be Van Der Walls, also frictional due to molecular shape.

I agree with you that Kinetic Interactions have a role - even in monatomic gases where the above forces are taken to be negligible, viscosity is fully explained by momentum exchange (Kinetic interactions)

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/Viscosity.htm

but this explanation only makes sense if the wall condition is already Vx =0

So much the same problem.- if the molecules are monatomic spheres, wall collsions are perfectly elastic, adhesion and cohesion are negligible

Then I am stuck with the same issue, what causes the no slip condition at the wall to set Vx = 0 at the wall-The Elastic Collison's -Dynamic Case Ideal Fluid suggests it wont be

The only way I can understand Vx=0 is by inelastic heat exchange-

Solid wall is absorbing original translational vector momentum as heat and providing from its heat , unrelated random translational momentum back to the fluid molecule.

This explanation also makes sense to me considering what @boneh3ad discusses above .
Any additional TP to the wall above thermal equilibrium/ static pressure - is treated as a heat gradient energy outflow from the fluid thru the wall

Last edited: Oct 4, 2016
13. Oct 5, 2016

### Staff: Mentor

I forgot to mention that this is all worked out in detail in Transport Phenomena, Bird, Stewart, and Lightfoot, Chapter 1.

14. Oct 5, 2016

### Timtam

Hi @Chestermiller thanks for the reference always appreciate additional reading materials

Unfortunately that text treats the no slip condition (Vx=0 at the surface) as the base assumption. , It then goes on to mathematically explain the impact of its presence (which i'm all fine with) but it does not explain its cause. Especially in gas where, for the reasons above, I see the traditional explanations fall down.

A blurb from an extract - http://link.springer.com/referenceworkentry/10.1007/978-3-540-30299-5_19

"The no-slip boundary condition at a solid–liquid interface is at the centre of our understanding of fluid mechanics. However, this condition is an assumption that cannot be derived from first principles."

Is that why I am having trouble finding an intuitive explanation of it ? It has been empirically proved so is fair to be assumed . But by that treatment, I shouldn't understand it to be understood ?

Last edited: Oct 5, 2016
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